The remainder when the square of any positive | vedclass

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(B) Let $a$ be any positive integer. Any integer $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$ for some integer $k ge 0$.We

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$2$

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(B) Let $a$ be any positive integer. Any integer $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$ for some integer $k ge 0$.We calculate $a^2 \pmod{6}$ for each case:$1$. If $a = 6k$,then $a^2 = 36k^2 = 6(6k^2)$,so $a^2 \equiv 0 \pmod{6}$.$2$. If $a = 6k+1$,then $a^2 = 36k^2 + 12k + 1 = 6(6k^2 + 2k) + 1$,so $a^2 \equiv 1 \pmod{6}$.$3$. If $a = 6k+2$,then $a^2 = 36k^2 + 24k + 4 = 6(6k^2 + 4k) + 4$,so $a^2 \equiv 4 \pmod{6}$.$4$. If $a = 6k+3$,then $a^2 = 36k^2 + 36k + 9 = 36k^2 + 36k + 6 + 3 = 6(6k^2 + 6k + 1) + 3$,so $a^2 \equiv 3 \pmod{6}$.$5$. If $a = 6k+4$,then $a^2 = 36k^2 + 48k + 16 = 36k^2 + 48k + 12 + 4 = 6(6k^2 + 8k + 2) + 4$,so $a^2 \equiv 4 \pmod{6}$.$6$. If $a = 6k+5$,then $a^2 = 36k^2 + 60k + 25 = 36k^2 + 60k + 24 + 1 = 6(6k^2 + 10k + 4) + 1$,so $a^2 \equiv 1 \pmod{6}$.The possible remainders are $0, 1, 3, 4$. The remainders $2$ and $5$ are not possible. Among the given options,$2$ is the correct answer.

The remainder when the square of any positive integer $a$ is divided by $6$ cannot be which of the following? $(1, 3, 4)$

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